Re: logical equivalence problem

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Mike et al. --

Back to the initial problem:

* Show that ~(p <=> q) is equivalent to (~q) <=> p

We can translate this into logical graphs by supposing that we
have to express everything in terms of negation and conjunction,
using parentheses for negation -- that is, "(x)" for "not x" --
and simple concatenation for conjunction -- "xyz" or "x y z"
for "x and y and z".

In this form of representation, for historical reasons called
the "existential interpretation" of logical graphs, we have
the following expressions for basic logical operations:

The disjunction "x or y" is written "((x)(y))".

This corresponds to the logical graph:

         x   y
         o   o
          \ /
           o
           |
           O

The disjunction "x or y or z" is written "((x)(y)(z))".

This corresponds to the logical graph:

         x y z
         o o o
          \|/
           o
           |
           O

Etc.

The implication "x => y" is written "(x (y)),
which can be read "not x without y" if that
helps to remember the form of expression.

This corresponds to the logical graph:

         y o
           |
         x o
           |
           O

Thus, the equivalence "x <=> y" has to be written somewhat
inefficiently as a conjunction of to and fro implications:
"(x (y))(y (x))".

This corresponds to the logical graph:

       y o   o x
         |   |
       x o   o y
          \ /
           O

Putting all the pieces together, the problem given
amounts to proving the following equation, expressed
in parse string and logical graph forms, respectively:

* Show that ~(p <=> q) is equivalent to (~q) <=> p

       q o   o p           q o
         |   |               |
       p o   o q             o   o p
          \ /                |   |
           o               p o   o--o q
           |                  \ /
           O         =         O

( (p (q)) (q (p)) ) = (p ( (q) )) ((p)(q))

No kidding ...

Jon Awbrey

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